Draw the structure of the product formed when 2-methylbutanal is treated with cold aqueous base.

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21-04-2018
Chemistry

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Draw the structure of the product formed when 2-methylbutanal is treated with cold aqueous base.

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transitionstate transitionstate · Answer 1 · 2018-04-30T18:17:36+02:00

When 2-methylbutnal is treated with base it looses its proton at alpha position as this proton is mildly acidic in nature and results in the formation of Enolate. The enolatae formed attacks on the carbonyl group of next aldehyde and forms a product called Aldol. Such reactions are called as Aldol Condensation. And those aldol condensation reactions in which the enolate and carbonyl group belongs to same carbonyl molecule like in our case, are called as Self Aldol Condensation reaction.

nikitarahangdaleVT nikitarahangdaleVT · Answer 2 · 2022-04-22T08:33:34+02:00

The product formed formed when 2-methylbutanal is treated with cold aqueous base is 2 ethyl 3 hydroxy 2,3,4 trimethyl hexanal.

What is aldol condensation?

Aldol condensation is a reaction in which dehydration takes place to the formation of β-hydroxy aldehyde or β-hydroxy ketone as a reaction between the enolate ion and carbonyl group in the presence of acid or base.

Steps involved in the aldol condensation reaction of 2-methylbutanal are:

Firstly alpha hydrogen of 2-methylbutanal is looses from the molecule in the presence of cold aqueous base for the formation of enolate ion.
Then this formed enolate ion will attack on the carbonyl group of the second molecule of the 2-methylbutanal for the formation of product i.e. 2 ethyl 3 hydroxy 2,3,4 trimethyl hexanal.

Image of the product is attached below and this type of condensation is also known as self condensation.

Hence, structure of the product is attached below.

To know more about aldol condensation, visit the below link:

https://brainly.com/question/9960404